Q.A bag contains 4 red and 4 black balls, another bag contains 2 red and 6 black balls. One of the two bags is selected at random and a ball is drawn from the bag which is found to be red. Find the probability that the ball is drawn from the first bag.

Showing the 12 most recent of 45 on this concept.

1. COMEDK 2025Set 2025-A1 markMCQ
   Q.Three bags contain a number of red and white balls are as follows. Bag I: 3 red balls Bag II: 2 red balls and 1 white ball Bag III: 3 White balls The probability that bag $i$ will be chosen and a ball is selected from it is $\frac{i}{6}, i=1,2,3$. If a white ball is selected, what is the probablity that it came from Bag III (A) $\frac{9}{11}$ (B) $\frac{2}{11}$ (C) $0$ (D) $\frac{1}{11}$

   ›Reveal solutionSolution

   This is a classic Bayes’ theorem problem: we are given prior probabilities for choosing each bag and the conditional probabilities of drawing a white ball from each bag; after observing a white ball, we want the posterior probability that it came from Bag III. The answer is $\frac{9}{11}$.

   We start by noting that the problem gives us three bags with different compositions, and a rule for choosing a bag: the probability of picking bag $i$ is $\frac{i}{6}$. Then, from that bag, we pick a ball uniformly at random. We are told that a white ball was selected, and we need the probability that it came from Bag III. This is a textbook case for Bayes’ theorem, which lets us “reverse” conditional probabilities: we know $P(\text{white} \mid \text{bag } i)$, and we want $P(\text{bag } i \mid \text{white})$.

   Let’s work through it step by step.

   1. Define events and write down given probabilities. Let $B_i$ be the event that bag $i$ is chosen ($i=1,2,3$). We are told:

   $P(B_1) = \frac{1}{6},\quad P(B_2) = \frac{2}{6},\quad P(B_3) = \frac{3}{6}.$

   Let $W$ be the event that a white ball is drawn.

   From the bag compositions:

   • Bag I: 3 red, 0 white → $P(W \mid B_1) = 0$.
   • Bag II: 2 red, 1 white → $P(W \mid B_2) = \frac{1}{3}$.
   • Bag III: 0 red, 3 white → $P(W \mid B_3) = 1$.

   2. Compute the total probability of drawing a white ball. By the law of total probability:

   $P(W) = P(B_1)P(W\mid B_1) + P(B_2)P(W\mid B_2) + P(B_3)P(W\mid B_3).$

   Substitute:

   $P(W) = \frac{1}{6}\cdot 0 + \frac{2}{6}\cdot\frac{1}{3} + \frac{3}{6}\cdot 1 = 0 + \frac{2}{18} + \frac{3}{6} = \frac{1}{9} + \frac{1}{2}.$

   Get a common denominator (18):

   $P(W) = \frac{2}{18} + \frac{9}{18} = \frac{11}{18}.$

   3. Apply Bayes’ theorem to find $P(B_3 \mid W)$. Bayes’ theorem says:

   $P(B_3 \mid W) = \frac{P(B_3)\,P(W\mid B_3)}{P(W)}.$

   Plug in the numbers:

   $P(B_3 \mid W) = \frac{\frac{3}{6} \cdot 1}{\frac{11}{18}} = \frac{\frac{1}{2}}{\frac{11}{18}} = \frac{1}{2} \cdot \frac{18}{11} = \frac{9}{11}.$

   Tip

   Notice that Bag I contributes zero probability of a white ball, so it can be ignored entirely in the numerator — it only affects the denominator by adding zero. This often simplifies Bayes’ calculations: only bags that can produce the observed outcome matter.

   Watch out

   A common mistake is to forget that the prior probabilities $\frac{i}{6}$ are not equal — they favor higher-numbered bags. If you mistakenly treated all bags as equally likely, you would get $\frac{1}{2}$ instead of $\frac{9}{11}$.

   Thus, given that a white ball was drawn, the probability it came from Bag III is $\frac{9}{11}$.

   ✓Final answer

   The correct option is (A).

   ANSWER: A

2. COMEDK 2024Set 2024-E1 markMCQ
   Q.Suppose we have three cards identical in form except that both sides of the first card are coloured red, both sides of the second are coloured black, and one side of the third card is coloured red and the other side is coloured black. The three cards are mixed and a card is picked randomly. If the upper side of the chosen card is coloured red, what is the probability that the other side is coloured black. (A) $\frac{1}{6}$ (B) $\frac{1}{2}$ (C) 0 (D) $\frac{1}{3}$

   ›Reveal solutionSolution

   This is a classic conditional probability problem (often called the "three cards" or "Bertrand's box" variant). The key is that seeing a red side updates the probability space to only the red sides, and among those, only one belongs to the mixed card. The answer is 1/3.

   We are asked: given that the visible side is red, what is the probability that the other side is black? This is not simply "one of the two remaining cards has a black other side" because the cards are not equally likely once we condition on the observation.

   ---

   Concept and Intuition

   The pitfall is to think: "We see red, so the card is either the all-red or the mixed card. That's two possibilities, so the chance is 1/2." But this ignores that the all-red card has two red sides, while the mixed card has only one red side. When we pick a card at random and then look at a random side, each of the six sides is equally likely to be the one we see. Seeing red eliminates the three black sides, leaving only the three red sides. Among those three red sides, two belong to the all-red card and only one belongs to the mixed card. So the probability that the other side is black is 1/3.

   ---

   Step-by-step reasoning

   1. Label the sides.

      Let the cards be:

      • Card A: both sides red (sides $R_1, R_2$)
      • Card B: both sides black (sides $B_1, B_2$)
      • Card C: one red, one black (sides $R_3, B_3$)

   2. Count equally likely outcomes.

      When we pick a card uniformly at random and then look at a random side, there are $3 \times 2 = 6$ equally likely side-views. Each of the six sides has probability $1/6$ of being the one we see.

   3. Condition on seeing a red side.

      The red sides are: $R_1, R_2$ (from card A) and $R_3$ (from card C). That's 3 red sides. So the conditional space has 3 equally likely possibilities.

   4. Identify which of these have the other side black.

      Only the red side $R_3$ (from card C) has a black other side. The other two red sides ($R_1, R_2$) have red on the other side.

   5. Compute the conditional probability.

   $P(\text{other side black} \mid \text{top side red}) = \frac{\text{number of favorable red sides}}{\text{total number of red sides}} = \frac{1}{3}.$

   Watch out

   The common mistake is to forget that the all-red card contributes two red sides to the sample space, making it twice as likely as the mixed card once we condition on seeing red.

   Tip

   A neat way to think: "Probability = (number of red sides with black opposite) / (total number of red sides)." This avoids the card-counting trap entirely.

   ---

   ✓Final answer

   The correct option is (D).

   ANSWER: D

3. COMEDK 2023Set 2023-E1 markMCQ
   Q.Bag A contains 3 white and 2 red balls. Bag B contains only 1 white ball. A fair coin is tossed. If head appears then 1 ball is drawn at random from bag A and put into bag B. However if tail appears then 2 balls are drawn at random from bag A and put into bag B. Now one ball is drawn at random from bag B. Given that the drawn ball from B is white, the probability that head appeared on the coin is (A) $\frac{23}{30}$ (B) $\frac{12}{23}$ (C) $\frac{11}{23}$ (D) $\frac{19}{30}$

   ›Reveal solutionSolution

   Computing $P(\text{white drawn}\mid H)=\frac45$ and $P(\text{white}\mid T)=\frac{11}{15}$, Bayes' theorem gives $P(H\mid\text{white})=\frac{12}{23}$.

   Head (transfer 1 ball from A$(3W,2R)$ to B$(1W)$, then draw from B's 2 balls):

   $P(\text{white})=\tfrac35(1)+\tfrac25\left(\tfrac12\right)=\tfrac35+\tfrac15=\tfrac45.$

   Tail (transfer 2 balls from A to B$(1W)$, then draw from B's 3 balls):

   $P(WW)=\tfrac{3}{10}\Rightarrow P(\text{white})=1,\quad P(WR)=\tfrac{6}{10}\Rightarrow \tfrac23,\quad P(RR)=\tfrac{1}{10}\Rightarrow \tfrac13.$

   $P(\text{white}\mid T)=\tfrac{3}{10}+\tfrac{6}{10}\cdot\tfrac23+\tfrac{1}{10}\cdot\tfrac13=\tfrac{3}{10}+\tfrac{4}{10}+\tfrac{1}{30}=\tfrac{22}{30}=\tfrac{11}{15}.$

   With $P(H)=P(T)=\tfrac12$:

   $P(H\mid\text{white})=\frac{\tfrac12\cdot\tfrac45}{\tfrac12\cdot\tfrac45+\tfrac12\cdot\tfrac{11}{15}}=\frac{\tfrac45}{\tfrac45+\tfrac{11}{15}}=\frac{\tfrac{12}{15}}{\tfrac{23}{15}}=\frac{12}{23}.$

   ✓Final answer

   The correct option is (B) — $\frac{12}{23}$

4. COMEDK 2021Set 2021-B1 markMCQ
   Q.An urn contains 5 red and 5 black coloured balls. A ball is picked at random, its colour noted and then is put back into the urn. Now, additional two balls of the same colour are put into the urn and then a ball is drawn at random. What is the probability that the ball now chosen is red. (A) 7/24 (B) 1/2 (C) 5/8 (D) 2/3

   ›Reveal solutionSolution

   By symmetry the second-draw red probability averages to $\tfrac12$.

   Start: 5 red, 5 black (10 total). First ball is drawn, replaced, then 2 more of the same colour are added, so the urn always has 12 balls before the second draw.

   • First ball red ($P=\tfrac{5}{10}=\tfrac12$): urn now 7R, 5B ⇒ $P(\text{red})=\tfrac{7}{12}$.
   • First ball black ($P=\tfrac12$): urn now 5R, 7B ⇒ $P(\text{red})=\tfrac{5}{12}$.

   Total probability:

   $\frac12\cdot\frac{7}{12}+\frac12\cdot\frac{5}{12}=\frac{7+5}{24}=\frac{12}{24}=\frac12.$

   ✓Final answer

   The correct option is (B) — 1/2

5. COMEDK 2025Set 2025-E1 markMCQ
   Q.A pot contains 5 red and 2 green balls. A ball is drawn at random from this pot. If a drawn ball is green, then a red ball is added to the pot. If a drawn ball is red, then a green ball is added to the pot, while the original ball drawn is not replaced in the pot. Now a second ball is drawn at random from the pot, what is the probability that the second ball drawn is a red ball? (A) $\frac{12}{49}$ (B) $\frac{32}{49}$ (C) $\frac{3}{7}$ (D) $\frac{27}{49}$

   ›Reveal solutionSolution

   Condition on the first draw: $P(\text{2nd red}) = \tfrac{2}{7}\cdot\tfrac{6}{7} + \tfrac{5}{7}\cdot\tfrac{4}{7} = \tfrac{32}{49}$.

   Start: $5$ red, $2$ green, total $7$. The first ball is not replaced, and a ball of the other colour is added.

   Case 1 — first ball is green. $P = \dfrac{2}{7}$.

   Remove that green ($5R,1G$), then add a red $\Rightarrow 6R,1G$ (total $7$).

   $P(\text{2nd red}\mid \text{green first}) = \frac{6}{7}$

   Case 2 — first ball is red. $P = \dfrac{5}{7}$.

   Remove that red ($4R,2G$), then add a green $\Rightarrow 4R,3G$ (total $7$).

   $P(\text{2nd red}\mid \text{red first}) = \frac{4}{7}$

   Total probability:

   $P(\text{2nd red}) = \frac{2}{7}\cdot\frac{6}{7} + \frac{5}{7}\cdot\frac{4}{7} = \frac{12}{49} + \frac{20}{49} = \frac{32}{49}$

   ✓Final answer

   $P(\text{second ball red}) = \dfrac{32}{49}$ — option (B).

6. COMEDK 2024Set 2024-M1 markMCQ
   Q.An urn contains 2 white and 2 black balls. A ball is drawn at random. If it is white it is not replaced into the urn. Otherwise it is replaced along with another ball of the same colour. The process is repeated. The probability that the third ball drawn is black is (A) $\frac{17}{30}$ (B) $\frac{37}{60}$ (C) $\frac{31}{60}$ (D) $\frac{23}{30}$

   ›Reveal solutionSolution

   Conditioning on the first two draws (white -> not replaced; black -> replaced plus one extra black) and using total probability gives $P(\text{3rd black}) = \tfrac{23}{30}$ — option (D).

   Rules. Start with 2 white, 2 black (total 4). Drawing white removes it (whites $-1$, total $-1$). Drawing black puts it back and adds one more black (blacks $+1$, total $+1$).

   First two draws and the composition just before the 3rd draw

   • WW: $\tfrac24\cdot\tfrac13 = \tfrac16$; urn becomes W0, B2 (total 2) $\Rightarrow P(\text{black}) = 1$.
   • WB: $\tfrac24\cdot\tfrac23 = \tfrac13$; urn becomes W1, B3 (total 4) $\Rightarrow P(\text{black}) = \tfrac34$.
   • BW: $\tfrac24\cdot\tfrac25 = \tfrac15$; urn becomes W1, B3 (total 4) $\Rightarrow P(\text{black}) = \tfrac34$.
   • BB: $\tfrac24\cdot\tfrac35 = \tfrac{3}{10}$; urn becomes W2, B4 (total 6) $\Rightarrow P(\text{black}) = \tfrac23$.

   (The four path probabilities sum to $\tfrac16+\tfrac13+\tfrac15+\tfrac{3}{10} = 1$.)

   Total probability

   $P(\text{3rd black}) = \tfrac16(1) + \tfrac13\cdot\tfrac34 + \tfrac15\cdot\tfrac34 + \tfrac{3}{10}\cdot\tfrac23$

   $= \frac{10}{60} + \frac{15}{60} + \frac{9}{60} + \frac{12}{60} = \frac{46}{60} = \frac{23}{30}$

   ✓Final answer

   $P(\text{third ball is black}) = \dfrac{23}{30}$ — option (D).

   ANSWER: D

7. COMEDK 2026Set 2026-A1 markMCQ
   Q.Vishnu has two jars of marbles, Jar A and Jar B. Jar A contains 3 yellow marbles and 2 green marbles. Jar B contains 4 yellow marbles and 3 green marbles. Vishnu flips a fair coin. If it lands heads, he picks two marbles at random without replacement from Jar A. If it lands tails, he picks two marbles at random with replacement from Jar B. Given that Vishnu picked one yellow and one green marble, what is the probability that they came from Jar B? (A) $\frac{21}{41}$ (B) $\frac{49}{89}$ (C) $\frac{40}{89}$ (D) $\frac{20}{41}$

   ›Reveal solutionSolution

   $P(E\mid A)=\tfrac35$ (without replacement), $P(E\mid B)=\tfrac{24}{49}$ (with replacement); Bayes gives $P(B\mid E)=\dfrac{40}{89}$ — option (C).

   Likelihoods of drawing one yellow and one green ($E$).

   Jar A (3 yellow, 2 green; two draws without replacement):

   $P(E\mid A)=\frac{\binom{3}{1}\binom{2}{1}}{\binom{5}{2}}=\frac{6}{10}=\frac{3}{5}.$

   Jar B (4 yellow, 3 green; two draws with replacement):

   $P(E\mid B)=2\cdot\frac{4}{7}\cdot\frac{3}{7}=\frac{24}{49}.$

   Bayes' theorem with $P(A)=P(B)=\tfrac12$ (the $\tfrac12$ cancels):

   $P(B\mid E)=\frac{P(E\mid B)}{P(E\mid A)+P(E\mid B)}=\frac{\frac{24}{49}}{\frac{3}{5}+\frac{24}{49}}.$

   Combine the denominator over $245$:

   $\frac{3}{5}+\frac{24}{49}=\frac{147+120}{245}=\frac{267}{245}.$

   $P(B\mid E)=\frac{24}{49}\cdot\frac{245}{267}=\frac{24\cdot 5}{267}=\frac{120}{267}=\frac{40}{89}.$

   ✓Final answer

   $P(B\mid E)=\dfrac{40}{89}$, which is option (C).

8. COMEDK 2025Set 2025-E1 markMCQ
   Q.Two numbers are selected at random from integers 1 to 9 . If their sum is even, what is the probability that both the numbers are odd? (A) $\frac{4}{9}$ (B) $\frac{5}{8}$ (C) $\frac{1}{6}$ (D) $\frac{2}{3}$

   ›Reveal solutionSolution

   This is a conditional probability problem: given that the sum of two numbers from 1–9 is even, we want the probability both are odd. The answer is 5/8, option (B).

   We are selecting two numbers from 1 to 9 without replacement (since "selected at random" from distinct integers usually implies no repetition). The sum is even only if both numbers are odd or both are even. So the condition restricts us to those pairs. The question asks: among those pairs with an even sum, what fraction consists of two odd numbers?

   ---

   1. Count total possible pairs (without replacement)

   From 1 to 9, there are 9 numbers. The number of ways to choose any two distinct numbers is

   $\binom{9}{2} = 36.$

   2. Count pairs with an even sum

   A sum is even when both numbers have the same parity.

   • Odd numbers from 1 to 9: 1, 3, 5, 7, 9 → 5 odds.

     Number of odd–odd pairs: $\binom{5}{2} = 10$.

   • Even numbers from 1 to 9: 2, 4, 6, 8 → 4 evens.

     Number of even–even pairs: $\binom{4}{2} = 6$.

   So total pairs with an even sum:

   $10 + 6 = 16.$

   3. Apply conditional probability

   We want

   $P(\text{both odd} \mid \text{sum even}) = \frac{\text{Number of odd–odd pairs}}{\text{Number of even-sum pairs}} = \frac{10}{16} = \frac{5}{8}.$

   Tip

   A common mistake is to treat this as an unconditional probability and compute $\frac{10}{36}$, which gives $\frac{5}{18}$ — not even among the options. The condition "given sum is even" changes the denominator from 36 to 16.

   Watch out

   Another pitfall: forgetting that selection is without replacement. If replacement were allowed, the counts would differ, but here the problem implies distinct integers.

   ---

   ✓Final answer

   The correct option is (B).

   ANSWER: B

9. KCET 2021Set A-11 markMCQ
   Q.A car manufacturing factory has two plants X and Y. Plant X manufactures 70% of cars and plant Y manufactures 30% of cars. 80% of cars at plant X and 90% of cars at plant Y are rated as standard quality. A car is chosen at random and is found to be of standard quality. The probability that it has come from plant X is (A) $\dfrac{56}{73}$ (B) $\dfrac{56}{84}$ (C) $\dfrac{56}{83}$ (D) $\dfrac{56}{79}$

   ›Reveal solutionSolution

   This is a reverse-probability question (effect → cause), so use Bayes' theorem with the total probability of a standard-quality car in the denominator.

   Step 1 — Define the events.

   • $X$: the car came from plant X — $P(X) = 0.70$
   • $Y$: the car came from plant Y — $P(Y) = 0.30$
   • $S$: the car is of standard quality

   The conditional (likelihood) data given:

   $P(S\mid X) = 0.80, \qquad P(S\mid Y) = 0.90$

   Note $X$ and $Y$ are mutually exclusive and exhaustive ($0.7 + 0.3 = 1$), which is exactly what Bayes' theorem needs.

   Step 2 — Why Bayes and not simple conditioning.

   We are told the effect (the chosen car is standard) and asked for the probability of the cause (it came from X). That inversion — $P(X\mid S)$ from $P(S\mid X)$ — is precisely Bayes' theorem:

   $P(X \mid S) = \frac{P(X)\,P(S\mid X)}{P(X)\,P(S\mid X) + P(Y)\,P(S\mid Y)}$

   Step 3 — Compute the numerator.

   $P(X)P(S\mid X) = 0.70 \times 0.80 = 0.56$

   Step 4 — Compute the denominator (total probability of a standard car).

   $P(S) = 0.70\times0.80 + 0.30\times0.90 = 0.56 + 0.27 = 0.83$

   Step 5 — Divide.

   $P(X\mid S) = \frac{0.56}{0.83} = \frac{56}{83}$

   Step 6 — Sanity check. The posterior $56/83 \approx 0.675$ is a little below the prior $0.70$ — which makes sense, because plant Y has the higher standard-quality rate ($90\% > 80\%$), so learning the car is standard shifts a little belief towards Y. ✓

   (The distractors 73, 84, 79 are what you get by mis-adding $0.56 + 0.27$; the denominator must be the total probability $0.83$.)

   ✓Final answer

   The correct option is (C) $\frac{56}{83}$.

   ANSWER: C

10. KCET 2025Set A-11 markMCQ
    Q.Meera visits only one of the two temples A and B in her locality. Probability that she visits temple A is $\dfrac{2}{5}$. If she visits temple A, $\dfrac{1}{3}$ is the probability that she meets her friend, whereas it is $\dfrac{2}{7}$ if she visits temple B. Meera met her friend at one of the two temples. The probability that she met her at temple B is (A) $\dfrac{7}{16}$ (B) $\dfrac{5}{16}$ (C) $\dfrac{3}{16}$ (D) $\dfrac{9}{16}$

    ›Reveal solutionSolution

    The friend has already been met (the effect); we want the probability of the cause (temple B) — that reversal of conditioning is exactly Bayes' theorem.

    Step 1 — Name the events.

    Let $A$ = "Meera visits temple A", $B$ = "Meera visits temple B", $F$ = "she meets her friend".

    She visits only one of the two temples, so $A$ and $B$ are mutually exclusive and exhaustive:

    $P(A)=\frac25\quad\Longrightarrow\quad P(B)=1-\frac25=\frac35 .$

    Given: $P(F\mid A)=\dfrac13$, $P(F\mid B)=\dfrac27$.

    Step 2 — Why Bayes.

    We are told the outcome ($F$ happened) and asked for the probability of a cause ($B$). Bayes' theorem inverts the conditioning:

    $P(B\mid F)=\frac{P(B)\,P(F\mid B)}{P(A)\,P(F\mid A)+P(B)\,P(F\mid B)} .$

    The denominator is $P(F)$ by the law of total probability — the friend can be met on either branch.

    Step 3 — Compute the two branch probabilities.

    $P(A\cap F)=\frac25\times\frac13=\frac{2}{15},\qquad P(B\cap F)=\frac35\times\frac27=\frac{6}{35} .$

    Step 4 — Total probability of meeting the friend.

    LCM of 15 and 35 is 105:

    $\frac{2}{15}=\frac{14}{105},\qquad \frac{6}{35}=\frac{18}{105},$

    $P(F)=\frac{14}{105}+\frac{18}{105}=\frac{32}{105}.$

    Step 5 — Apply Bayes.

    $P(B\mid F)=\frac{18/105}{32/105}=\frac{18}{32}=\frac{9}{16}.$

    (Check: $P(A\mid F)=\frac{14}{32}=\frac{7}{16}$, and $\frac{9}{16}+\frac{7}{16}=1$ ✓. Note option (A) $\frac{7}{16}$ is the trap — it is the probability for temple A.)

    ✓Final answer

    The correct option is (D) — $\dfrac{9}{16}$.

    ANSWER: D

11. COMEDK 2025Set 2025-A1 markMCQ
    Q.A bag contains $(n+1)$ coins. It is known that one of these coins has a head on both sides, whereas the other coins are fair. One of these coins is selected at random and tossed. If the probability that the toss results in heads is $\frac{7}{12}$, then the value of $n$ is : (A) 5 (B) 3 (C) 2 (D) 4

    ›Reveal solutionSolution

    The key idea is to treat the coin selection as a two‑case partition (two‑headed coin vs. fair coins) and apply the law of total probability. Solving the resulting equation gives $n = 5$, so the correct option is (A).

    ---

    Concept and intuition

    We have a mixed bag: one trick coin that always lands heads, and $n$ fair coins that land heads with probability $\frac12$. When we pick a coin at random and toss it, the overall chance of heads is a weighted average of the two cases. The weight for the trick coin is $\frac{1}{n+1}$ (since there are $n+1$ coins total), and for any fair coin it’s $\frac{n}{n+1}$. The problem gives that overall probability as $\frac{7}{12}$, so we set up an equation and solve for $n$.

    ---

    Step‑by‑step solution

    1. Define the events Let $T$ be the event that the two‑headed coin is selected, and $F$ the event that a fair coin is selected. Since selection is random,

    $P(T) = \frac{1}{n+1}, \qquad P(F) = \frac{n}{n+1}.$

    2. Conditional probabilities for heads If the trick coin is chosen, heads is certain:

    $P(\text{heads} \mid T) = 1.$

    If a fair coin is chosen, the chance of heads is $\frac12$:

    $P(\text{heads} \mid F) = \frac12.$

    3. Apply the law of total probability The overall probability of heads is

    $P(\text{heads}) = P(T) \cdot P(\text{heads}\mid T) + P(F) \cdot P(\text{heads}\mid F).$

    Substituting the values:

    $P(\text{heads}) = \frac{1}{n+1} \cdot 1 + \frac{n}{n+1} \cdot \frac12.$

    4. Simplify the expression

    $P(\text{heads}) = \frac{1}{n+1} + \frac{n}{2(n+1)} = \frac{2 + n}{2(n+1)}.$

    5. Set equal to the given probability The problem states this equals $\frac{7}{12}$:

    $\frac{2 + n}{2(n+1)} = \frac{7}{12}.$

    6. Solve for $n$ Cross‑multiply:

    $12(2 + n) = 7 \cdot 2(n+1) \quad\Rightarrow\quad 24 + 12n = 14n + 14.$

    Bring terms together:

    $24 - 14 = 14n - 12n \quad\Rightarrow\quad 10 = 2n.$

    Hence

    $n = 5.$

    Tip

    A quick sanity check: with $n=5$, there are 6 coins total. The trick coin contributes $\frac16$ to the overall heads probability, and the five fair coins contribute $\frac56 \cdot \frac12 = \frac{5}{12}$. Sum: $\frac{1}{6} + \frac{5}{12} = \frac{2}{12} + \frac{5}{12} = \frac{7}{12}$. Perfect.

    Watch out

    A common mistake is forgetting that the denominator is $n+1$, not $n$. Always count all coins, including the trick coin.

    ---

    ✓Final answer

    The correct option is (A).

    ANSWER: A

12. COMEDK 2026Set 2026-M1 markMCQ
    Q.Advika chooses one of three scarves every morning: Red, Blue, or Green. The probability she chooses Red is $20 \%$. The probability she chooses Blue is twice the probability of choosing Red. On the remaining days she wears a Green scarf. Once a scarf is chosen, she decides whether to wear a Hat (H) and Sunglasses (S). These choices are independent of each other but depend on the scarf colour: $\begin{array}{|l|l|l|} \hline \text { Scarf colour } & \mathbf{P ( H )} & \mathbf{P ( S )} \\ \hline \text { Red } & 0.5 & 0.8 \\ \hline \text { Blue } & 0.4 & 0.5 \\ \hline \text { Green } & 0.1 & 0.5 \\ \hline \end{array}$ Advika is spotted outdoors wearing both a Hat and Sunglasses. What is the probability that she is wearing the Red scarf? (A) $\frac{13}{313}$ (B) $\frac{8}{21}$ (C) $\frac{4}{9}$ (D) $\frac{8}{13}$

    ›Reveal solutionSolution

    Bayes' theorem on scarf colour given that both a hat and sunglasses are worn. Priors $P(R)=0.2,\ P(B)=0.4,\ P(G)=0.4$; likelihoods $P(H\cap S\mid\text{colour})=P(H)\,P(S)$. The posterior $P(R\mid H\cap S)=\frac{4}{9}$ — option (C).

    Concept. Hat and sunglasses are independent given the scarf, so $P(H\cap S\mid\text{colour})=P(H\mid\text{colour})\cdot P(S\mid\text{colour})$. Bayes' theorem then reverses the conditioning to give the probability of the scarf colour from the observed accessories.

    Step 1 — Priors.

    $P(R)=20\%=0.2,\qquad P(B)=2P(R)=0.4,\qquad P(G)=1-0.2-0.4=0.4.$

    Step 2 — Likelihood of wearing both accessories for each colour.

    $P(H\cap S\mid R)=0.5\times0.8=0.40,$

    $P(H\cap S\mid B)=0.4\times0.5=0.20,$

    $P(H\cap S\mid G)=0.1\times0.5=0.05.$

    Step 3 — Total probability of both accessories (denominator).

    $P(H\cap S)=(0.2)(0.40)+(0.4)(0.20)+(0.4)(0.05)=0.08+0.08+0.02=0.18.$

    Step 4 — Posterior for Red.

    $P(R\mid H\cap S)=\frac{P(R)\,P(H\cap S\mid R)}{P(H\cap S)}=\frac{0.08}{0.18}=\frac{8}{18}=\frac{4}{9}.$

    ✓Final answer

    $P(\text{Red}\mid H\cap S)=\dfrac{4}{9}$ — option (C).

    ANSWER: C